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Laws of Motion question

2015 · Q137
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Laws of Motion question

2015 · Q137

NEETPhysicsLaws of MotionMCQ+4 / −1
A block A of mass m1 rests on a horizontal table. A light string connected to it passes over a frictionless pully at the edge of table and from its other end another block B of mass m2 is suspended. The coefficient of kinetic friction between the block and the table is μ\muμk. When the block A is sliding on the table, the tension in the string is
  1. A
    m1m2(1+μk)g(m1+m2){{{m_1}{m_2}(1 + {\mu _k})g} \over {({m_1} + {m_2})}}(m1​+m2​)m1​m2​(1+μk​)g​
  2. B
    m1m2(1−μk)g(m1+m2){{{m_1}{m_2}(1 - {\mu _k})g} \over {({m_1} + {m_2})}}(m1​+m2​)m1​m2​(1−μk​)g​
  3. C
    (m2+μkm1)g(m1+m2){{\left( {{m_2} + {\mu _k}{m_1}} \right)g} \over {\left( {{m_1} + {m_2}} \right)}}(m1​+m2​)(m2​+μk​m1​)g​
  4. D
    (m2−μkm1)g(m1+m2){{\left( {{m_2} - {\mu _k}{m_1}} \right)g} \over {\left( {{m_1} + {m_2}} \right)}}(m1​+m2​)(m2​−μk​m1​)g​
View written solutionFree

Correct answer: A

For the motion of both the blocks
m1a = T – μk{\mu _k}μk​m1g
m2g – T = m2a

AIPMT 2015 Cancelled Paper Physics - Laws of Motion Question 47 English Explanation
a=m2g−μkm1gm1+m2a = {{{m_2}g - {\mu _k}{m_1}g} \over {{m_1} + {m_2}}}a=m1​+m2​m2​g−μk​m1​g​

m2g−T=(m2)(m2g−μkm1gm1+m2){m_2}g - T = \left( {{m_2}} \right)\left( {{{{m_2}g - {\mu _k}{m_1}g} \over {{m_1} + {m_2}}}} \right)m2​g−T=(m2​)(m1​+m2​m2​g−μk​m1​g​)

solving we get tension in the string

T=m1m2(1+μk)gm1+m2T = {{{m_1}{m_2}\left( {1 + {\mu _k}} \right)g} \over {{m_1} + {m_2}}}T=m1​+m2​m1​m2​(1+μk​)g​

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